Is 2i less than 3i? (Part 3: An inequality that almost works)

In yesterday’s post, I demonstrated that there is no subset of the complex numbers which satisfies the following four axioms:

If , then

If , then .

For every , either or , but not both.

However, it’s instructive (and fun) to try to construct such a set. One way of attempting this is defining

This set leads to the lexicographic ordering of the complex numbers: if and , where , we say that if

I used the symbol because, as we’ll see, satisfies some but not all of the usual properties of an inequality. This ordering is sometimes called the “dictionary” order because the numbers are ordered like the words in a dictionary… the real parts are compared first, and then (if that’s a tie) the imaginary parts are compared. See Wikipedia and Mathworld for more information.

In any case, defining in this way satisfies three of the four order axioms.

Suppose . It’s straightforward to show that . Let and , where . Then , and so .

Case 1: If , then clearly .

Case 2: If , that’s only possible if and . But since , that means that and . Therefore, . Since , we again conclude that .

Suppose , where . Then or . We now show that, no matter what, or , but not both.

Case 1: If , then , and so but .

Case 2: If , then , and so but .

Case 3: If , then since . Also, if , then , so that and .

Subcase 3A: If , then , and so but .

Subcase 3B: If , then , and so but .

By definition, .

However, the fourth property fails. By definition, . However, .

Because this definition of satisfies three of the four order axioms, the relation satisfies some but not all of the theorems stated in the first post of this series. For example, if and , then . Also, if and , then .

I’ll leave it to the interested reader to determine which of the theorems are true, and which are false (and have counterexamples).